cho a=\(\dfrac{-1+\sqrt{2}}{2}\) và b=\(\dfrac{-1-\sqrt{2}}{2}\)
Tính a7+b7
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1
14 tháng 4 2023
a: Khi x=25 thì \(A=\dfrac{7\cdot5-2}{5-2}=\dfrac{33}{3}=11\)
b: P=A*B
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{2}{\sqrt{x}-1}-\dfrac{4\sqrt{x}}{x-1}\right)\cdot\dfrac{7\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+2\sqrt{x}+2-4\sqrt{x}}{x-1}\cdot\dfrac{7\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\dfrac{x-3\sqrt{x}+2}{x-1}\cdot\dfrac{7\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\cdot\left(7\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{7\sqrt{x}-2}{\sqrt{x}+1}\)
31 tháng 8 2023
1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)
2: \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
3: A/B>3/2
=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)
=>\(-\sqrt{x}+2>0\)
=>-căn x>-2
=>căn x<2
=>0<x<4
H9
HT.Phong (9A5)
CTVHS
31 tháng 8 2023
1) Thay x=64 vào A ta có:
\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)
2) \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
3) Ta có:
\(\dfrac{A}{B}>\dfrac{3}{2}\) khi
\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{2-\sqrt{x}}{2\sqrt{x}}>0\)
Mà: \(2\sqrt{x}\ge0\forall x\)
\(\Leftrightarrow2-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
Kết hợp với đk:
\(0< x< 4\)
20 tháng 8 2021
b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
1 tháng 9 2021
Bài 2:
Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
18 tháng 10 2021
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)
4 tháng 3 2021
b) Ta có: \(4x^2+x-5=0\)
\(\Leftrightarrow4x^2-4x+5x-5=0\)
\(\Leftrightarrow4x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{5}{4}\left(loại\right)\end{matrix}\right.\)
Thay x=1 vào biểu thức \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\), ta được:
\(B=\dfrac{\sqrt{1}-1}{\sqrt{1}}=0\)
Vậy: Khi \(4x^2+x-5=0\) thì B=0
Ta có: \(a+b=\dfrac{-1+\sqrt{2}}{2}+\dfrac{-1-\sqrt{2}}{2}=-1\)
\(ab=\dfrac{-1+\sqrt{2}}{2}.\dfrac{-1-\sqrt{2}}{2}=\dfrac{-1}{4}\)
\(\left(a+b\right)^2=a^2+b^2+2ab=1\)
\(\Rightarrow a^2+b^2+2.\dfrac{-1}{4}=1\)\(\Rightarrow a^2+b^2=\dfrac{3}{2}\) (1)
\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)=-1\)
\(\Rightarrow a^3+b^3+3.\dfrac{-1}{4}.\left(-1\right)=-1\)\(\Rightarrow a^3+b^3=\dfrac{-7}{4}\) (2)
Từ (1) và (2) suy ra \(\left(a^2+b^2\right)\left(a^3+b^3\right)=\dfrac{3}{2}.\dfrac{-7}{4}=\dfrac{-21}{8}\)
\(\Rightarrow a^5+a^3b^2+a^2b^3+b^5=\dfrac{-21}{8}\)
\(\Rightarrow a^5+b^5+a^2b^2\left(a+b\right)=\dfrac{-21}{8}\)
\(\Rightarrow a^5+b^5+\dfrac{1}{16}.\left(-1\right)=\dfrac{-21}{8}\)\(\Rightarrow a^5+b^5=\dfrac{-41}{16}\) (3)
Từ (1) và (3) suy ra \(\left(a^2+b^2\right)\left(a^5+b^5\right)=\dfrac{3}{2}.\dfrac{-41}{16}=\dfrac{-123}{32}\)
\(\Rightarrow a^7+a^5b^2+a^2b^5+b^7=\dfrac{-123}{32}\)
\(\Rightarrow a^7+b^7+a^2b^2\left(a^3+b^3\right)=\dfrac{-123}{32}\)
\(\Rightarrow a^7+b^7+\dfrac{1}{16}.\dfrac{-7}{4}=\dfrac{-123}{32}\)\(\Rightarrow a^7+b^7=\dfrac{-239}{64}\)
Vậy \(a^7+b^7=\dfrac{-239}{64}\)